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1. (a) 2 x 2 = 4 mod 4 = 0, but 0 is not in {1,2,3}. => not closed under mult mod 4. not a group
(b) to show some structure is a group, you need to check all the group axioms
first check if the defined mult is closed?
for all a in {1,2,3,4}, a x 1 = 1 x a = a in {1,2,3,4}
2 x 2 = 4 in {1,2,3,4}
3 x 3 = 9 mod 5 = 4 in {1,2,3,4}
4 x 4 = 16 mod 5 = 1 in {1,2,3,4}
2 x 3 = 3 x 2 = 6 mod 5 = 1 in {1,2,3,4}
2 x 4 = 4 x 2 = 8 mod 5 = 3 in {1,2,3,4}
3 x 4 = 4 x 3 = 12 mod 5 = 2 in {1,2,3,4}
=> closed under mult mod 5. [closed under * means for all a,b in G, a*b in G]
your way of showing closed under mult may be better, but i don't know it
the second step is to check all the group axioms
(i) is the operation associative?
(ii) is there an identity element?
(iii) given a in {1,2,3,4}, is there an inverse for a?
the answer is yes for all these questions [you need to show]. now you can conclude that it is a group.
2. to show GL(2,R) is a non-abelian group, you just need to come up with an example AB =/ BA, where A, B in GL(2,R)
for A to be in GL(2,R), det(A) must not equal 0. so yes you need to check for the determinant
here is one example [you can come up with many other examples]: let
[1 2] = A
[0 1]
[1 0] = B
[2 1]
det(A) = -1 so A in GL(2,R). [A is invertible]. det(B) = -1 so B in GL(2,R)
you can check that
[5 2] = AB
[2 1]
[1 2] = BA
[2 5]
so AB =/ BA; thus GL(2,R) is non-abelian
3. again you have to check everything. first check if it closed under mult (which you already did)
next you need to check all the group axioms. let's call the set described above G (set of 3x3 matrices of the form above)
(i) associative? yes because matrix multiplication is associative
(ii) identity? yes, just the identity matrix I
[1 0 0]
[0 1 0] = I
[0 0 1]
(here a = b = c = 0, I is in G)
(iii) inverse?
you are looking for the inverse of
[1 a b]
[0 1 c] = A
[0 0 1]
you can check that the determinant is not 0 (detA = 1), so the inverse exists. there is a way to calculate the inverse of a 3x3 matrix [you can try that]. and check to see if the inverse is in G
the easy way is to use the matrix you had
1 a b][1 a' b']
[0 1 c][0 1 c']
[0 0 1][0 0 1]
=
[1 a+a' b'+ac'+b]
[0 001 0000c'+c]
[0 000 0000001]
you want the last matrix above to be I, which means
a+a' = 0 => a' = -a
c'+c = 0 => c' = -c
b'+ac'+b = 0 => b' = -ac' - b = ac - b [because c' = -c]
so the inverse of A should be
[1 -a ac-b]
[0 01 -c] = B
[0 00 01]
note that that B is in G. now check to show that AB = I = BA. thus B = A^{-1}
that is, you are done!
_________________ The word analysis comes from the Greek ανάλυσις, which means "dissolving into pieces."http://www.facebook.com/home.php?ref=ho ... 2306439078
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